Every line of 'python sort array of integers' code snippets is scanned for vulnerabilities by our powerful machine learning engine that combs millions of open source libraries, ensuring your Python code is secure.

By copying the Snyk Code Snippets you agree to

this disclaimer

145 def counting_sort_ints(arr): 146 """ 147 for array [7,1,5,2,2] len = 5, range of values from 0 = 0 to 7 148 the algorithm is 149 O(len) to count (assuming that finding max and min is O(1)) 150 O(range) for cumulative sum 151 O(len) to copy back 152 153 so O(2n + k) = O(n) where k is the range of items and assumed to be less than the input size. 154 155 if the range is big (like in big_arr), the complexity is dominated by k 156 157 However in application, k usually small. 158 """ 159 c1, c2, c3 = 0, 0, 0 # Use to look at the counts. 160 161 # Set up 162 max_number = max(arr) 163 count = [0] * (max_number+1) # is the array of "buckets" which starts at 0 and goes to max+1 164 output = [0] * len(arr) 165 166 # Count occurrences of each number in arr and put it in 'bucket' in count. 167 for number in arr: # the item at index number of count += 1 to found occurrence of that number 168 count[number] += 1 169 c1 += 1 170 171 # Cumulative sum of occurrences. 172 for i in range(1, len(count)): # cumulative sum 173 count[i] += count[i-1] 174 c2 += 1 175 176 # Put into output stably. 177 for j in range(len(arr)-1, -1, -1): # work backwards to keep stable 178 output_idx = count[arr[j]] - 1 # -1 as output len = arr len 179 output[output_idx] = arr[j] # put in right place in output 180 count[arr[j]] -= 1 # decrement value in count 181 c3 += 1 182 print("first loop counting: " + str(c1) + "\nsecond loop summing: " + str(c2) + "\nthird loop copying: " + str(c3)) 183 return output

Important

Secure your code as it's written. Use Snyk Code to scan source code in minutes – no build needed – and fix issues immediately. Enable Snyk Code

16 def CombSort(array): 17 18 # The inc starts from the highest value 19 inc = len(array) 20 21 # This boolean tracks whether a swap occured in the current iteration 22 swap = True 23 24 # if inc is 1 and no swaps are happening, the array has been sorted 25 # if either of them are true, trudge forward because work needs to be done 26 while not inc == 1 or swap == True: 27 28 # Get a new value for the inc 29 inc = RecalibrateInc(inc) 30 31 # Set swap to False 32 swap = False 33 34 # Do the bubb.. bubb.. CombSort 35 # len(array) - inc because len(array) - inc + inc onwards, no elements exist 36 for i in range(0, len(array) - inc): 37 38 if array[i] > array[i + inc]: 39 40 # swap the array elements 41 array[i], array[i + inc] = array[i + inc], array[i] 42 43 swap = True

115 def counting_sort(arr): 116 c1, c2, c3 = 0, 0, 0 117 118 # set up 119 max_number = max(arr) 120 count = [0] * (max_number+1) # is the array of "buckets" which starts at 0 and goes to max+1 121 output = [0] * len(arr) 122 123 # count occurrences of each number in arr and put it in 'bucket' in count 124 for number in arr: # the item at index number of count += 1 to found occurrence of that number 125 count[number] += 1 126 127 c1 += 1 128 129 # cumulative sum of occurrences 130 for i in range(1, len(count)): # cumulative sum 131 count[i] += count[i-1] 132 133 c2 += 1 134 135 # put into output stably 136 for j in range(len(arr)-1, -1, -1): # work backwards to keep stable 137 output_idx = count[arr[j]] - 1 # -1 as output len = arr len 138 output[output_idx] = arr[j] # put in right place in output 139 count[arr[j]] -= 1 # decrement value in count 140 141 print(output) 142 c3 += 1 143 144 print("first loop: " + str(c1) + "\nsecond loop: " + str(c2) + "\nthird loop: " + str(c3)) 145 """ 146 for array [7,1,5,2,2] len = 5, range of values from 0 = 0 to 7 147 the algorithm is 148 O(len) to count (and find max?) 149 O(range) for cumulative sum 150 O(len) to copy back 151 152 so O(3n + k) = O(n) 153 154 if the range is big (like in big_arr), the complexity is dominated by k 155 156 however in application, k usually small 157 """ 158 return output

102 def sorttest(A): 103 bubblesort(A)

38 def sort(a): 39 mergeSort(a,0,len(a)-1)

113 def sort(packed, ref, reverse=True): 114 """ 115 Sort a series of packed list, according to a ref list. 116 Also return the original index before the sort. 117 """ 118 assert (isinstance(packed, tuple) or isinstance(packed, list)) and isinstance(ref, list) 119 packed = [ref] + [range(len(ref))] + list(packed) 120 sorted_packed = [list(t) for t in zip(*sorted(zip(*packed), reverse=reverse))] 121 return tuple(sorted_packed[1:])

7 def BitonicSort(direction, arr): 8 9 ''' 10 11 The obvious condition that we have over here is that is the length of the array is less than or equal to 1, then it makes no sense to call the BitonicMerge function on the array as it contains only one element. 12 13 ''' 14 15 if len(arr) <= 1: 16 17 return arr 18 19 else: 20 21 ''' 22 23 When you split the array to form a Bitonic Sequence, you want the first half to be in an increasing order and the second half to be in a decreasing order. The booleans True and False respresent increasing and decreasing order of values respectively and this will become more obvious in the BitonicCompare functions. 24 25 ''' 26 27 first = BitonicSort(True, arr[:len(arr) // 2]) 28 second = BitonicSort(False, arr[len(arr) // 2:]) 29 30 ''' 31 32 This is the only line that might be a little confusing and the reason for this is very simple. When we looked at the example, we executed the increasing and decreasing sorting of the array to form Bitonic Sequences simultaneously. As it turns out, we are not following the same procedures here. In fact, these operations are being done at very different times in the execution of the array but don't let that confuse you. This is the part in the process where you move to the next stage and increase the size of the Bitonic Sequence in the array. 33 34 ''' 35 36 return BitonicMerge(direction, first + second)